Proof
Suppose $\sqrt{2} = \frac{p_o}{q_o}$, where $q_o$ is as small as possible.
We have $q_o < p_o < 2q_o$ (from $1 < \sqrt{2} < 2$, which follows from $1 < 2 < 4$)
let $q_1 = p_o - q_o$, and $p_1 = 2q_o - p_o$ (where both $p_1$ and $q_1$ are natural numbers)
Then, $\frac{p_1}{q_1} =\frac{2q_o-p_o}{p_o-q_o}$ $=\frac{(2q_o-\sqrt{2}qo)}{\sqrt{2}q_o - q_o}$ $=\frac{\sqrt{2}q_o(\sqrt{2} - 1)}{q_o(\sqrt{2} - 1)} $ $= \sqrt{2}$
but $q_1 < q_o$, so contradiction.
Questions:
How does $q_o < p_o <2q_o$ follow from $1 < \sqrt{2} < 2$? I know it can also be written as $\frac{p_o}{\sqrt{2}} < p_o < \sqrt{2}p_o$, but I don't know how this is related to $1 < \sqrt{2} < 2$.
How does $q_1 < q_o$ prove $\sqrt{2} is irrational? Which premise does it contradict?
For your question 1, we take the known fact $1<\sqrt{2}<2$ and multiply everything by $q_0$, giving us $q_0 < q_0 \sqrt{2} < 2q_0$. Then we note that by hypothesis $q_0 \sqrt{2} = p_0$.
For your question 2, we supposed (in your very first line) that $q_0$ was the smallest (positive) denominator of any fraction equalling $\sqrt{2}$. You then proceeded to produce another representation $\sqrt{2}=p_1/q_1$ with a smaller denominator. That's the contradiction.